Homework Help: Proof that f(z) is in no point analytic

1. Sep 4, 2014

jennyjones

1. The problem statement, all variables and given/known data

I have to proof for a homework assignment that f(z) = z^(n)(z*)^(-m) where n, m ≥ 1 (are natural numbers) is in no point analytic.

2. Relevant equations

binomial function

3. The attempt at a solution

I found this wikipedia page, were they state that:
a complex-valued function ƒ of a complex variable z is said to be analytic at a if in some open disk centered at a it can be expanded as a convergent power series.

for wich i think i need to use the binomial function.

But i'm not sure how to do this for the given function f(z)

Last edited: Sep 4, 2014
2. Sep 4, 2014

jennyjones

(z*) is the complex conjugate

If i look at the Lim (z->0) z/z* i would say that it does not exist because if i look let z approach 0 on the real axis i get 1, and if i let z approach 0 on the imaginary axis i get -1.

but now i think i have to look at the lim(z->0) z^n/(z*)^m and am not sure how to approach the powers.

thanx

Last edited: Sep 4, 2014
3. Sep 4, 2014

vela

Staff Emeritus
4. Sep 4, 2014

jennyjones

So the function is analytic in a domain if it is differentiable at all points of the domain.

5. Sep 4, 2014

Xiuh

There is yet another criterion based on the so called Wirtinger derivatives

see: http://wcherry.math.unt.edu/math5410/wirtinger.pdf

Roughly, if $\frac{\partial f(z,\overline{z})}{\partial \overline{z}} \neq 0$, then $f$ does not satisfy the Cauchy-Riemann equations, and thus cannot be analytic.

Edit: Here $\overline{z}$ is the conjugate of $z$

Last edited: Sep 4, 2014